Show that the Fundamental Theorem of Algebra is true for the quadratic polynomial −2x2+4x−9=0

by using the quadratic formula. Which of the following statements accurately describes the solution set?(1 point)
Responses

There are two non-real solutions.
There are two non-real solutions.

There are two identical solutions.
There are two identical solutions.

There are two rational solutions.
There are two rational solutions.

There are two irrational solutions.

1 answer

To determine the nature of the solutions for the quadratic polynomial \(-2x^2 + 4x - 9 = 0\), we'll use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our specific polynomial, the coefficients are:

  • \(a = -2\)
  • \(b = 4\)
  • \(c = -9\)

Now, we'll substitute these values into the quadratic formula.

  1. Calculate the discriminant \(b^2 - 4ac\): \[ b^2 = 4^2 = 16 \] \[ 4ac = 4 \cdot (-2) \cdot (-9) = 72 \] \[ b^2 - 4ac = 16 - 72 = -56 \]

  2. Since the discriminant is negative (\(-56\)), we will have two non-real solutions.

  3. Now we can calculate the solutions using the quadratic formula: \[ x = \frac{-4 \pm \sqrt{-56}}{2(-2)} \] Simplifying further: \[ x = \frac{-4 \pm \sqrt{56}i}{-4} = \frac{4 \mp \sqrt{56}i}{4} \] \[ x = 1 \mp \frac{\sqrt{56}}{4} i \]

  4. Therefore, the solutions are: \[ x = 1 + \frac{\sqrt{56}}{4} i \quad \text{and} \quad x = 1 - \frac{\sqrt{56}}{4} i \]

As the solutions are complex (non-real), the correct statement about the solution set is:

There are two non-real solutions.